See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When propene (CH3-CH=CH2) reacts with Br2 under hv (light/photochemical conditions) at low concentration, the reaction proceeds via a free radical mechanism rather than electrophilic addition. Step 1 - Identify the reaction conditions: Br2/hv indicates photochemical (free radical) bromination. Low concentration of Br2 further confirms this is not an electrophilic addition (which would require darkness or no radical initiator). Step 2 - Free radical bromination mechanism: Under photochemical conditions, Br2 undergoes homolytic cleavage to give Br• radicals. These radicals abstract a hydrogen atom from the substrate. Step 3 - Allylic bromination: In propene, the allylic C-H bond (at the CH3 group adjacent to the double bond, i.e., C3 position) is weakest and most susceptible to radical abstraction because the resulting allylic radical is stabilized by resonance. The allylic position in propene is the -CH3 group (C1 in CH3-CH=CH2 numbering from the methyl end, or the terminal carbon). Step 4 - Product formation: The allylic radical (•CH2-CH=CH2) reacts with Br2 to give allyl bromide: H2C=CH-CH2-Br (3-bromoprop-1-ene / allyl bromide). The double bond remains intact. Step 5 - Why other options fail: - Option (a): CH3-CH(Br)-CH2-Br is a vicinal dibromide, product of ionic electrophilic addition of Br2 (without light, in dark conditions). - Option (c): CH3-C(Br)=CH2 would require addition across the double bond or a different mechanism; not the allylic radical product. - Option (d): Br-CH2-CH2-CH2-Br is a 1,3-dibromide, which is not the product of monosubstitution under low Br2 concentration. Therefore, the correct answer is B.